Answer key

Answers

1. Kinetic energy is ½mv² = ½(3)(4)² = ½(3)(16) = 24 J.
2. The gain is mgy = (2.5)(9.8)(1.6) = 39.2 J.
3. Elastic energy is ½kx² = ½(120)(0.20)² = ½(120)(0.04) = 2.4 J.
4. The push is along the motion, so W = Fd = (50)(2.5) = 125 J.
5. The upward force is perpendicular to the horizontal walk, so cos 90° = 0. That force does 0 J of work. The box’s height does not change, so its gravitational energy does not change either.
6. Power is W/t = 1800/12 = 150 W.
7. Efficiency is useful output divided by input: 600/800 = 0.75, or 75%.
8. On a smooth ramp, mgh = ½mv², so the mass cancels. v = √(2gh) = √(2·9.8·0.45) = √8.82 = 3.0 m/s (2.97 m/s before rounding).
9. Energy at the bottom: mgh − 2.0 = 4.41 − 2.0 = 2.41 J = ½mv². v = √(2·2.41) = 2.2 m/s.
10. Useful work is the same for both: mgy = (25)(9.8)(2.0) = 490 J. Power of A is 490/5 = 98 W. Power of B is 490/20 = 24.5 W. A is more powerful because it does that work sooner, not because it does more of it.